Question

Two point charges of 1.0 microcoulomb and -0.25 microcoulomb are placed in air at a distance of 0.40 metre from each other find out at what point on the line joining the two charges should third charge be placed so that no force act upon it

Answer 1

Answer:

-1.2 m

Explanation:

Let the charges be Q1, Q2 and Q3. Let the distance between Q1 and Q3 be x and the distance between Q3 and Q2 be (x + 0.40)

The force between the charges are :

kQ1Q3/x² = kQ2Q3/(x + 0.4)²

We divide both sides by kQ3 to get :

Q1/x² = Q2/(x + 0.40)²

1 × 10^-6/x² = (- 0.25 × 10^-6) / (x + 0.40)²

(1 × 10^-6)/(-0.25 × 10^-6) = x²/(x + 0.40)²

4 = x²/(x + 0.40)²

Getting square root of both sides.

2 = x/(x + 0.40)

2(x + 0.40) = x

2x + 0.80 = x

2x - x = - 0.80

x = - 0.80

From the two charges the position of the third charge is :

= 0.80 + 0.40 = 1.2 m

The position of charge 3 is - 1.2 m